
A battery is charged at a potential of 15 V for 8 hours when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 hours. The mean terminal voltage during discharge is 14 V. The “Watt hour” efficiency of the battery is:
A. 80%
B. 90%
C. 87.5%
D. 82.5%
Answer
484.2k+ views
Hint: Efficiency is generally expressed in percentage and it is the ratio of useful energy to the energy supplied to the battery.
The product of power i.e., product of potential difference and current and time gives us the work.
Complete step by step answer:
When a current of I ampere flows through a battery for a time t second and a source of potential difference V volt is connected across its ends. The amount of electrical energy supplied by the source is obtained by the definition of potential difference i.e., when Q coulomb of electric charge flows through a potential difference of V volt, the work done W is given by
$W = QV$
$\implies W = \left( {It} \right)V\left[ {\because Q = I \times t} \right]$
$\implies W = VIt$ joule or Watt-hour (The energy is also defined as a product of power and time. Unit of power is watt and time is hour)
Hence, the energy supplied to the battery when it is charged at a potential difference of 15V for 8 hours and the current flowing through it is 10A is ${W_1}$.
${W_1} = 15 \times 10 \times 8$
$\implies {W_1} = 1200J$
The energy produced when the battery is discharged at terminal voltage 14V and the current supplied by it is 5A for the time is 15 hours is ${W_2}$.
${W_2} = 14 \times 5 \times 15$
$\implies {W_2} = 1050J$
The efficiency of the battery is obtained by dividing energy produced by energy supplied and it is expressed as percentage.
$Efficiency = \dfrac{{{W_1}}}{{{W_2}}} \times 100\% $
$\implies Efficiency = \dfrac{{1050}}{{1200}} \times 100\% $
$\therefore Efficiency = 87.5\% $
So, the correct answer is “Option C”.
Note:
The efficiency in watt-hour means the unit of energy is watt-hour which is the same as joule.
The product of power and time is numerically equal to work or energy.
The efficiency of the battery is obtained by dividing energy produced by energy supplied and it is expressed as percentage.
The product of power i.e., product of potential difference and current and time gives us the work.
Complete step by step answer:
When a current of I ampere flows through a battery for a time t second and a source of potential difference V volt is connected across its ends. The amount of electrical energy supplied by the source is obtained by the definition of potential difference i.e., when Q coulomb of electric charge flows through a potential difference of V volt, the work done W is given by
$W = QV$
$\implies W = \left( {It} \right)V\left[ {\because Q = I \times t} \right]$
$\implies W = VIt$ joule or Watt-hour (The energy is also defined as a product of power and time. Unit of power is watt and time is hour)
Hence, the energy supplied to the battery when it is charged at a potential difference of 15V for 8 hours and the current flowing through it is 10A is ${W_1}$.
${W_1} = 15 \times 10 \times 8$
$\implies {W_1} = 1200J$
The energy produced when the battery is discharged at terminal voltage 14V and the current supplied by it is 5A for the time is 15 hours is ${W_2}$.
${W_2} = 14 \times 5 \times 15$
$\implies {W_2} = 1050J$
The efficiency of the battery is obtained by dividing energy produced by energy supplied and it is expressed as percentage.
$Efficiency = \dfrac{{{W_1}}}{{{W_2}}} \times 100\% $
$\implies Efficiency = \dfrac{{1050}}{{1200}} \times 100\% $
$\therefore Efficiency = 87.5\% $
So, the correct answer is “Option C”.
Note:
The efficiency in watt-hour means the unit of energy is watt-hour which is the same as joule.
The product of power and time is numerically equal to work or energy.
The efficiency of the battery is obtained by dividing energy produced by energy supplied and it is expressed as percentage.
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